Galois groups of finite abelian extensions of ℚ

Theorem.

Let G be a finite abelian group with |G|>1. Then there exist infinitely many number fieldsK with K/Q Galois and Gal⁡(K/Q)≅G.

Proof.

This will first be proven for G cyclic.

Let |G|=n. By Dirichlet’s theorem on primes in arithmetic progressions, there exists a prime p with p≡1⁢mod⁡n. Let ζp denote a pthroot of unity. Let L=ℚ⁢(ζp). Then L/ℚ is Galois with Gal⁡(L/ℚ) cyclic of order (http://planetmath.org/OrderGroup) p-1. Since n divides p-1, there exists a subgroupH of Gal⁡(L/ℚ) such that |H|=p-1n. Since Gal⁡(L/ℚ) is cyclic, it is abelian, and H is a normal subgroup of Gal⁡(L/ℚ). Let K=LH, the subfield of L fixed (http://planetmath.org/FixedField) by H. Then K/ℚ is Galois with Gal⁡(K/ℚ) cyclic of order n. Thus, Gal⁡(K/ℚ)≅G.

Let p and q be distinct primes with p≡1⁢mod⁡n and q≡1⁢mod⁡n. Then there exist subfields K1 and K2 of ℚ⁢(ζp) and ℚ⁢(ζq), respectively, such that Gal⁡(K1/ℚ)≅G and Gal⁡(K2/ℚ)≅G. Note that K1∩K2=ℚ since ℚ⊆K1∩K2⊆ℚ⁢(ζp)∩ℚ⁢(ζq)=ℚ. Thus, K1≠K2. Therefore, for every prime p with p≡1⁢mod⁡n, there exists a distinct number field K such that K/ℚ is Galois and Gal⁡(K/ℚ)≅G. The theorem in the cyclic case follows from using the full of Dirichlet’s theorem on primes in arithmetic progressions: There exist infinitely many primes p with p≡1⁢mod⁡n.

The general case follows immediately from the above , the fundamental theorem of finite abelian groups (http://planetmath.org/FundamentalTheoremOfFinitelyGeneratedAbelianGroups), and a theorem regarding the Galois group of the compositum of two Galois extensions.
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